Calculate the pH of a 0.200 M NaCH3CO2 Solution
This premium calculator determines the pH of sodium acetate solutions by modeling the base hydrolysis of acetate ion in water. Enter the concentration, choose how to provide the equilibrium constant, and instantly see the pH, pOH, Kb, hydroxide concentration, and a concentration-versus-pH chart.
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pH Trend for Sodium Acetate Solutions
The chart below shows how pH changes as sodium acetate concentration changes while the acetic acid equilibrium constant is held fixed. Your selected concentration is highlighted within the dataset.
How to calculate the pH of a 0.200 M NaCH3CO2 solution
Sodium acetate, written as NaCH3CO2 or more commonly NaC2H3O2, is the salt formed from a strong base and a weak acid. The strong base part is sodium hydroxide, and the weak acid part is acetic acid. Because the cation Na+ is essentially neutral in water and the acetate ion CH3CO2– is the conjugate base of a weak acid, the solution becomes basic. That means a 0.200 M sodium acetate solution has a pH greater than 7 at 25 C.
The key idea is hydrolysis. Once sodium acetate dissolves, it dissociates almost completely:
NaCH3CO2(aq) → Na+(aq) + CH3CO2–(aq)
The sodium ion does not significantly react with water, but acetate does:
CH3CO2– + H2O ⇌ CH3CO2H + OH–
This reaction generates hydroxide ions, so the solution is basic. The equilibrium constant for this reaction is Kb, the base dissociation constant for acetate. Since acetate is the conjugate base of acetic acid, its Kb is related to the acid dissociation constant Ka of acetic acid through:
Ka × Kb = Kw
At 25 C, Kw = 1.0 × 10-14. A commonly used value for acetic acid is Ka = 1.8 × 10-5, which corresponds to pKa ≈ 4.76. Therefore:
Kb = Kw / Ka = (1.0 × 10-14) / (1.8 × 10-5) = 5.56 × 10-10
Step-by-step setup using an ICE table
For a 0.200 M sodium acetate solution, the initial concentration of acetate is 0.200 M. Let x be the amount of acetate that hydrolyzes.
- Initial: [CH3CO2–] = 0.200, [CH3CO2H] = 0, [OH–] = 0
- Change: [CH3CO2–] = -x, [CH3CO2H] = +x, [OH–] = +x
- Equilibrium: [CH3CO2–] = 0.200 – x, [CH3CO2H] = x, [OH–] = x
Now apply the expression for Kb:
Kb = [CH3CO2H][OH–] / [CH3CO2–] = x2 / (0.200 – x)
Substitute the value of Kb:
5.56 × 10-10 = x2 / (0.200 – x)
Because Kb is very small relative to the concentration, the amount that reacts is tiny. That makes the common weak base approximation valid: 0.200 – x ≈ 0.200. Then:
x2 = (5.56 × 10-10)(0.200) = 1.11 × 10-10
x = [OH–] = √(1.11 × 10-10) = 1.05 × 10-5 M
Now calculate pOH:
pOH = -log(1.05 × 10-5) = 4.98
Finally:
pH = 14.00 – 4.98 = 9.02
Answer: pH ≈ 9.02 at 25 C
Why the answer is basic
Students often wonder why a salt can make water acidic or basic. The reason is that salts are not all neutral. A salt made from a strong acid and a strong base is usually neutral because neither ion significantly reacts with water. Sodium chloride is the classic example. But a salt made from a weak acid and a strong base often gives a basic solution because the anion is a conjugate base that pulls protons from water and creates hydroxide. Sodium acetate fits exactly into that category.
In contrast, if you had a salt made from a strong acid and a weak base, the cation could hydrolyze to produce hydronium and lower pH. This is why understanding the parent acid and base is the fastest conceptual method for predicting pH before calculating anything.
Shortcut formula for weak base salts
When the weak base approximation is valid, the hydroxide concentration for a weak base salt can be estimated from:
[OH–] ≈ √(KbC)
Then:
- Find Kb from Ka or pKa.
- Use the salt concentration as the initial base concentration.
- Compute [OH–].
- Find pOH and then pH.
For sodium acetate at 0.200 M:
- C = 0.200 M
- Kb = 5.56 × 10-10
- [OH–] ≈ √(5.56 × 10-10 × 0.200) = 1.05 × 10-5 M
- pH ≈ 9.02
This shortcut agrees with the exact quadratic solution to within a negligible amount for this case.
Exact versus approximate calculation
For many classroom and laboratory problems, the approximation method is accepted because the percent ionization is very small. Still, it is useful to compare the approximate and exact approaches so you can understand when each is appropriate.
| Method | Input values | Calculated [OH–] | pOH | pH at 25 C |
|---|---|---|---|---|
| Approximation | Kb = 5.56 × 10-10, C = 0.200 M | 1.054 × 10-5 M | 4.977 | 9.023 |
| Exact quadratic | x = [-Kb + √(Kb2 + 4KbC)] / 2 | 1.054 × 10-5 M | 4.977 | 9.023 |
| Difference | Same equilibrium constants | Less than 0.001% | Practically identical | Practically identical |
Because the approximation is so accurate here, most general chemistry solutions report the pH as 9.02. In advanced analytical work, however, exact solutions may be preferred, especially at very low concentrations or when multiple equilibria matter.
How concentration affects the pH of sodium acetate
As concentration increases, pH increases, but not in a linear way. Since hydroxide concentration depends on the square root of both Kb and concentration, a hundredfold change in concentration does not produce a hundredfold change in pH. Instead, pH changes gradually. This is why weak acid and weak base systems often show moderate pH shifts across wide concentration ranges.
| NaCH3CO2 concentration | Assumed Kb for acetate | Approximate [OH–] | Approximate pH at 25 C |
|---|---|---|---|
| 0.010 M | 5.56 × 10-10 | 2.36 × 10-6 M | 8.37 |
| 0.050 M | 5.56 × 10-10 | 5.27 × 10-6 M | 8.72 |
| 0.100 M | 5.56 × 10-10 | 7.46 × 10-6 M | 8.87 |
| 0.200 M | 5.56 × 10-10 | 1.05 × 10-5 M | 9.02 |
| 0.500 M | 5.56 × 10-10 | 1.67 × 10-5 M | 9.22 |
| 1.000 M | 5.56 × 10-10 | 2.36 × 10-5 M | 9.37 |
Common mistakes when solving this problem
1. Treating sodium acetate as a strong base
Sodium acetate is not the same as sodium hydroxide. The acetate ion is only a weak base, so the pH is not anywhere close to 13 or 14. The correct pH is only mildly basic, around 9.02.
2. Using Ka directly instead of converting to Kb
The reacting species is acetate, the conjugate base of acetic acid. That means you must either use Kb directly or convert from Ka using Kb = Kw / Ka. If you accidentally plug Ka into the weak base equation, your answer will be far off.
3. Forgetting that sodium ion is a spectator
Na+ comes from a strong base and does not contribute significantly to acidity or basicity under ordinary conditions. The acetate ion controls the pH.
4. Confusing pOH with pH
Since the calculation gives [OH–], the first logarithm gives pOH, not pH. You then convert using pH = pKw – pOH. At 25 C, pKw is 14.00.
5. Ignoring temperature assumptions
The familiar pH + pOH = 14 relationship is specifically tied to 25 C. At other temperatures, pKw changes. This calculator allows you to inspect how the result shifts when a different pKw is chosen.
When to use the Henderson-Hasselbalch equation and when not to
The Henderson-Hasselbalch equation is extremely useful for buffer systems that contain both a weak acid and its conjugate base in appreciable amounts. A pure sodium acetate solution is not initially a buffer in the usual prepared-mixture sense because you start mainly with the conjugate base. The direct hydrolysis method is cleaner and more rigorous for this problem. If acetic acid were also present in known concentration, then the buffer equation would become a convenient shortcut.
Practical significance of sodium acetate pH
Sodium acetate appears in food chemistry, buffer preparation, biochemical workflows, and analytical chemistry. Knowing that an aqueous sodium acetate solution is mildly basic helps with reaction planning, extraction conditions, and pH control. In biochemistry, acetate-based buffers can influence enzyme activity or the charge state of solutes. In general chemistry labs, sodium acetate is often used as a representative example of a basic salt because it illustrates the connection between weak acid conjugate bases and hydrolysis equilibria.
Authoritative references for equilibrium constants and pH concepts
- USGS: pH and Water
- NIST Chemistry WebBook: Acetic Acid Data
- University of Wisconsin Chemistry Resources
Final answer for the classic problem
To calculate the pH of a 0.200 M NaCH3CO2 solution, recognize that acetate is a weak base. Use the acetic acid constant to find Kb, write the hydrolysis equilibrium, and solve for hydroxide concentration. With Ka = 1.8 × 10-5 and pKw = 14.00 at 25 C, the calculation gives [OH–] ≈ 1.05 × 10-5 M, pOH ≈ 4.98, and therefore pH ≈ 9.02.
This means the solution is mildly basic, not strongly basic. If you need a fast memory aid, remember this pattern: salt of strong base + weak acid = basic solution. For sodium acetate, that rule predicts the correct direction of pH before you ever begin the arithmetic.